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ENGR12 Circuit Theory

Chapter 6

Capacitors and Inductors

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Capacitors and Inductors Chapter 6

6.1 Capacitors

6.2 Series and Parallel Capacitors

6.3 Inductors

6.4 Series and Parallel Inductors

The Capacitor

Capacitors are one of the fundamental passive

components. In its most basic form, it is composed

of two plates separated by a dielectric.

The ability to store charge is the definition of

capacitance.

Dielectric Conductors

Capacitance The Capacitor

The Capacitor

Dielectric

Plates

Leads

Electrons

BA

+

+

+

+

+

+

+

+

Initially uncharged

+

BA

+

+

+

+

+

+

+

Charging

+

BA

VS

+

+

+++++++++

Fully charged

BA

VS

+

+

+++++++++

Source removed

The charging

process…

A capacitor with stored charge can act as a temporary battery.

Capacitance is the ratio of charge to voltage

QC

V

Rearranging, the amount of charge on a

capacitor is determined by the size of the

capacitor (C) and the voltage (V).

Q CV

If a 22 mF capacitor is connected to

a 10 V source, the charge is 220 mC

Capacitance The Capacitor

Capacitance

An analogy:

Imagine you store rubber bands in a

bottle that is nearly full.

You could store more rubber bands

(like charge or Q) in a bigger bottle

(capacitance or C) or if you push

them in more (voltage or V). Thus,

Q CV

Capacitance The Capacitor

A capacitor stores energy in the form of an electric field

that is established by the opposite charges on the two

plates. The energy of a charged capacitor is given by the

equation

Capacitance

2

2

1CVW

where

W = the energy in joules

C = the capacitance in farads

V = the voltage in volts

Capacitance The Capacitor

The capacitance of a capacitor depends on

three physical characteristics.

128.85 10 F/m r AC

d

C is directly proportional to

and the plate area.

the relative dielectric constant

C is inversely proportional to

the distance between the plates

Capacitance Capacitance The Capacitor

128.85 10 F/m r AC

d

Find the capacitance of a 4.0 cm diameter

sensor immersed in oil if the plates are

separated by 0.25 mm.

The plate area is

The distance between the plates is

Capacitance

4.0 for oilr

3 2

12

3

4.0 1.26 10 m 8.85 10 F/m

0.25 10 mC

30.25 10 m

178 pF

2 2 3 2π 0.02 m 1.26 10 mA r

Capacitor types

Mica

Mica

Foil

Foil

Mica

Foil

Foil

Mica

Foil

Mica capacitors are small with high working voltage.

The working voltage is the voltage limit that cannot

be exceeded.

Capacitor types

Ceramic disk

Solder

Lead wire solderedto silver elec trode

Ceramicdielectric

Dipped phenolic coating

Silv er elec trodes deposited ontop and bottom of ceramic disk

Ceramic disks are small nonpolarized capacitors They

have relatively high capacitance due to high r.

Capacitor types

Capacitor types

Plastic Film

Lead wire

High-purityfoil electrodes

Plastic filmdielec tric

Outer wrap ofpolyester film

Capacitor section(alternate strips offilm dielectric andfoil electrodes)

Solder coated end

Plastic film capacitors are small and nonpolarized. They

have relatively high capacitance due to larger plate area.

Capacitor types Capacitor types

Capacitor types

Electrolytic (two types)

Symbol for any electrolytic capacitor

Al electrolytic

+

_

Ta electrolytic

Electrolytic capacitors have very high capacitance but

they are not as precise as other types and tend to have

more leakage current. Electrolytic types are polarized.

Capacitor types Capacitor types

Capacitor types

Variable

Variable capacitors typically have small capacitance

values and are usually adjusted manually.

A solid-state device that is used as a variable

capacitor is the varactor diode; it is adjusted with an

electrical signal.

Capacitor types Capacitor types

Capacitor labeling

Capacitors use several labeling methods. Small

capacitors values are frequently stamped on them such

as .001 or .01, which have units of microfarads.

+++

+

VT

TV

TT

47

MF

.022

Electrolytic capacitors have larger

values, so are read as mF. The unit is usually

stamped as mF, but some older ones may be

shown as MF or MMF).

A label such as 103 or 104 is read as 10x103

(10,000 pF) or 10x104 (100,000 pF)

respectively. (Third digit is the multiplier.)

Capacitor labeling

When values are marked as 330 or 6800, the

units are picofarads.

What is the value of

each capacitor? Both are 2200 pF.

222 2200

Capacitor labeling

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Capacitance, a Recap

• Capacitance occurs whenever electrical

conductors are separated by a dielectric,

or insulating material.

• Applying a voltage to the conductors

displaces the charge within the dielectric.

• Current does not actually flow through the

dielectric.

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Capacitor

• Circuit Symbol

• Component

designation (C)

• Units – Farads

– Usually μF or pF

• Reference directions

for voltage and

current

19

Voltage-Current Relationship

dvi C

dt

20

Observations

dvi C

dt

The voltage across a capacitor cannot change

instantaneously (the current would be infinite).

If the voltage across the terminals is constant, the

current will be zero.

(looks like an open circuit at DC).

Only a time-varying voltage can produce a

displacement current.

21

Example 1

The voltage across a 100-mF capacitor is

v(t) = 50 sin(120 t) V

Find the displacement current i flowing into capacitor as a function of time

Answer:

i(t) = 1.8850 cos(120 t)

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Express the voltage across the capacitor as

a function of the current

0 0

0

( )

( )

0

0

1

1

1( ) ( )

1( ) (0)

v t t

v t t

t

t

t

dvi C

dt

idt Cdv dv idtC

dx idC

v t id v tC

v t id vC

In English, please?

• Translation: to find the voltage at time t:

– take the known voltage at some start time t0

– add to it the integral from t0 to t of the current,

divided by C

23

What this looks like graphically

• At time 0 we measure

capacitor v = 5 V.

• A time varying current is then

applied to the capacitor

• Voltage at any time t is then

found as

v(t) =v(0)+ [area under i from 0 to t]/C

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25

Example 2

An initially uncharged 1-mF capacitor has the current shown below across it.

Calculate the voltage across it at t = 2 ms and t = 5 ms.

Answer:

v(2ms) = 100 mV

v(5ms) = 400 mV

V(2) = 0 + [area of Triangle]/.001F

V(5) = v(2) + [area of square]/.001F

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Power and Energy for the Capacitor

0

0 0

2

1( )

1

2

t

t

w v

dvp vi Cv

dt

p i id v tC

dw dvCv

dt dt

dw Cvdv

dx C ydy

w Cv

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Example 6.4

• A voltage pulse described as follows is applied across

the terminals of a 0.5μF capacitor:

• Derive the expressions for the capacitor a) current,

b) power, and c) energy.

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A) Derive the expressions for the capacitor current

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B) Derive the expressions for the capacitor power

30

C) Derive the expressions for the capacitor energy

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Energy is being stored whenever the

power is positive.

Energy is being delivered by the

capacitor whenever the power is

negative.

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6.2 Series and Parallel Capacitors (1)

• The equivalent capacitance of N parallel-connected capacitors is the sum of the individual capacitances.

Neq CCCC ...21

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6.2 Series and Parallel Capacitors (2)

• The equivalent capacitance of N series-connected capacitors is the reciprocal of the sum of the reciprocals of the individual capacitances.

Neq CCCC

1...

111

21

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Example 3

Find the equivalent capacitance seen at the terminals of the circuit in the circuit shown below:

Answer:

Ceq = 40mF

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6.3 Inductors (1)

• An inductor is a passive element designed to store energy in its magnetic field.

• An inductor consists of a coil of conducting wire.

Inductance is the property of a conductor to oppose

a change in current. The effect of inductance is

greatly magnified by winding a coil on a magnetic

material.

Inductance

Common symbols for inductors (coils) are

Air core Iron core Ferrite core Variable

Self Inductance

Self-inductance is usually just called inductance,

symbolized by L. Self-inductance is a measure of a coil’s

ability to establish an induced voltage as a result of a

change in its current. The induced voltage always opposes

the change in current, which is basically a statement of

Lenz’s law.

The unit of inductance is the henry (H). One henry is the

inductance of a coil when a current, changing at a rate of

one ampere per second, induces one volt across the coil.

Self Inductance

The induced voltage is given by the formula dt

diLv ind

What is the inductance if 37 mV is induced

across a coil if the current is changing at a rate of

680 mA/s?

54 mH ind

dtdi

vL

dt

diLv ind

Rearranging,

A/s 68.0

V 037.0

Factors affecting inductance

Four factors affect the amount of inductance for a

coil. The equation for the inductance of a coil is 2N A

Ll

m

where

L = inductance in henries

N = number of turns of wire

m = permeability in H/m (same as Wb/At-m)

l = coil length on meters

What is the inductance of a 2.0 cm long,

150 turn coil wrapped on an low carbon steel core

that is 0.50 cm diameter? The permeability of low

carbon steel is 2.5 x104 H/m (Wb/At-m).

22 5 2π π 0.0025 m 7.85 10 mA r

2N AL

l

m

22 mH

2 4 5 2150 t 2.5 10 Wb/At-m 7.85 10 m

0.020 m

Physical parameters affecting inductance

The inductance given by the equation in the

previous slide is for the ideal case. In practice,

inductors have winding resistance (RW) and

winding capacitance (CW). An equivalent circuit

for a practical inductor including these effects is:

LRW

CW

Recall Lenz’s law states,

When the current through a coil changes, an

induced voltage is created across the coil that

always opposes the change in current.

In a practical circuit, the current can change

because of a change in the load as shown in

the following circuit example...

Lenz’s law

Lenz’s law

A basic circuit to demonstrate Lenz’s law is shown.

Initially, the SW is open and there is a small

current in the circuit through L and R1.

R1

SW

R2VS

L

+

+

SW closes and immediately a voltage appears

across L that tends to oppose any change in current.

R1

SW

R2VS

L

+

+

+Initially, the meter

reads same current

as before the switch

was closed.

Lenz’s law

After a time, the current stabilizes at a higher level

(due to R2) as the voltage decays across the coil.

+

R1

SW

R2VS

L

+

Lenz’s law

Later, the meter

reads a higher

current because of

the load change.

Practical inductors

Inductors come in a variety of sizes. A few

common ones are shown here.

Encapsulated Torroid coil Variable

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Inductance

• Inductor

– A coil of wire wrapped around a supporting

core (magnetic or non-magnetic)

– The time-varying current in the wire produces

a time-varying magnetic field around the wire

– A voltage is induced in any conductor linked

by the magnetic field

– Inductance relates the induced voltage to the

current

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Inductor

• Circuit Symbol

• Component

designation (L)

• Units -- Henry(s)

– Usually mH or μH

• Reference directions

for voltage and

current

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Voltage-Current Relationship

div L

dt

The current across an inductor cannot

change instantaneously (the voltage would be

infinite).

If the current through the inductor is constant, the

voltage will be zero.

(looks like a short circuit at DC).

Only a time-varying current can produce an

induced voltage. 50

Observations

div L

dt

51

Current in an inductor in terms of the

Voltage across the inductor

0 0

0 0

( )

( )

0

1 1( ) ( ) (0)

i tt

t i t

t t

t t

div L

dt

divdt Ldi L dt

dt

vdt Ldi

vd L dx

i t vd i t vd iL L

Again, in English

• Translation: to find the current at time t:

– take the known current at some start time t0

– add to it the integral from t0 to t of the voltage,

divided by L

52

What this looks like graphically

• At time 0 we measure

inductor i = 5 A.

• A time varying voltage is then

applied to the inductor

• Current at any time t is then

found as

i(t) =i(0)+ [area under v from 0 to t]/L

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Power and Energy in an Inductor

0

0

0 0

2

1( )

1

2

t

t

w i

dip vi Li

dt

p vi v vd i tL

dw dip Li

dt dt

dw Lidi

dx L ydy

w Li

55

Example 5

The terminal voltage of a 2-H inductor is

v = 10(1-t) V

Find the current flowing through it at t = 4 s and the energy stored in it at t = 4 s.

Assume i(0) = 2 A.

Answer:

i(4s) = -18A

w(4s) = 320J

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Example 6

Determine vc, iL, and the energy stored in the capacitor and inductor in the circuit of circuit shown below under dc

conditions.

Answer:

iL = 3A

vC = 3V

wL = 1.125J

wC = 9J

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Example 6.1 – Sketch current, find max

using Wolfram Alpha

• First, sketch the current waveform.

• At what instant of time is the current maximum?

• Wolfram Alpha: Plot[ 10t*exp(-5t), t, 0, 1]

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5

5 5 5

10

10( 5 ) 10 (1 5 )

10 1 5 0 0.2

5

t

t t t

i te

dite e e t

dt

dit t s

dt

Plot(differentiate[ 10t*exp(-5t), t],t,0,1)

differentiate[ 10t*exp(-5t),t]

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Example 6.2 Using Derivative formula Given: current thru inductor

Find: voltage across inductor

5

5 5

10

(0.1)10 (1 5 ) (1 5 )

t

t t

i te

div L e t e t

dt

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Are the voltage and current at maximum at

the same time?

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At what instant of time does the voltage

change polarity?

The voltage changes polarity

when the current passes

through its maximum value

and the slope changes sign.

62

Is there ever an instantaneous change in

voltage across the inductor? If so, when?

Yes.

The voltage across the inductor changes

instantaneously at t=0.

Using wolframalpha.com to

differentiate • diff[ f(x), x ] takes derivative of some f(x) wrt x

• diff[10t*exp(-5t),t]

– Use * when needed to clarify

– Use exp( ) to raise e to some exponent

If you don’t like the plot range, hover mouse over letter A below plot,

Copy paste back into WA with new limits

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Example 6.3

• Plot i, v, p, and w for Example 6.3. Line

up the plots vertically to allow easy

assessment of each variable’s behavior.

• In what time interval is energy being

stored in the inductor?

• In what time interval is energy being

extracted from the inductor?

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Energy is being stored

when power is >0.

An increasing energy

curve indicates that

energy is being stored in

the inductor.

Energy is being extracted

when power is <0.

A decreasing energy curve

indicates that energy is

being extracted from the

inductor

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What is the maximum energy stored in the

inductor?

2 2

max

1 1(0.1)(0.736)

2 2

27.07

w Li

w mJ

70

Example 6.4 Using the integral formula Given: voltage across inductor

Find: current through inductor

• Sketch the voltage as a function of time.

• Find the inductor current as a function of time.

• Sketch the current as a function of time.

Integrating with Wolfram Alpha

• Basic indefinite integral:

– Integrate[x^2, {x}]

• Basic definite integral

– Integrate[x^2,{x,0,5}]

Integrating with Wolfram Alpha

• Our integrals must be done using “dummy

variable substitution” which is a definite

integral from t0 to t. In the last example,

t0=0

• In this example

– 0 + (1/0.1)*Integrate[20x*E^(-10x),{x,0,t}]

=

73

0

10

0

10

0

10 10

1( ) (0)

1( ) 20 0

0.1

( ) 200 (10 1)100

( ) 2(1 10 )

t

t

t

t t

i t vd iL

i t e d

ei t

i t te e

74

10 10( ) 2(1 10 )t ti t te e

Current approaches 2A as t ∞

75

6.4 Series and Parallel Inductors (1)

• The equivalent inductance of series-connected inductors is the sum of the individual inductances.

Neq LLLL ...21

76

6.4 Series and Parallel Inductors (2)

• The equivalent capacitance of parallel inductors is the reciprocal of the sum of the reciprocals of the individual inductances.

Neq LLLL

1...

111

21

77

Example 7

Calculate the equivalent inductance for the inductive ladder network in the circuit shown below:

Answer:

Leq = 25mH

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6.4 Series and Parallel Capacitors (4)

• Current and voltage relationship for R, L, C

+

+

+

Analog Integration http://ecee.colorado.edu/~mathys/ecen1400/labs/lab06/index.html

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A differential equation solver

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HANDOUTS

81

82

Example 1

The voltage across a 100-mF capacitor is

v(t) = 50 sin(120 t) V

Find the displacement current i flowing into capacitor as a function

of time

Answer:

i(t) = 1.8850 cos(120 t)

83

Example 2

An initially uncharged 1-mF capacitor has the current shown below across it.

Find the voltage across it at t = 2 ms and t = 5 ms.

Answer:

v(2ms) = 100 mV

v(5ms) = 400 mV

V(2) = 0 + [area of Triangle]/.001F

V(5) = v(2) + [area of square]/.001F

84

Example 3

Find the equivalent capacitance seen at the terminals of the circuit in the circuit shown below:

Answer:

Ceq = 40mF

85

Example 5

The terminal voltage of a 2-H inductor is

v = 10(1-t) V

Find the current flowing through it at t = 4 s and the energy stored in it at t = 4 s.

Assume i(0) = 2 A.

Answer:

i(4s) = -18A

w(4s) = 320J

86

Example 6

Determine vc, iL, and the energy stored in the capacitor and inductor in the circuit of circuit shown

below under dc conditions.

Answer:

iL = 3A

vC = 3V

wL = 1.125J

wC = 9J

87

Example 7

Calculate the equivalent inductance for the inductive ladder network in the circuit shown below:

Answer:

Leq = 25mH